Solve -8r^2+26r-15

$$- 8 r ^ { 2 } + 26 r - 15$$

Factor

$$\left(3-4r\right)\left(2r-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -8r^{2}+ar+br-15. To find a and b, set up a system to be solved.

$$a+b=26 ab=-8\left(-15\right)=120$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 120.

$$1,120 2,60 3,40 4,30 5,24 6,20 8,15 10,12$$

Calculate the sum for each pair.

$$1+120=121 2+60=62 3+40=43 4+30=34 5+24=29 6+20=26 8+15=23 10+12=22$$

The solution is the pair that gives sum 26.

$$a=20 b=6$$

Rewrite -8r^{2}+26r-15 as \left(-8r^{2}+20r\right)+\left(6r-15\right).

$$\left(-8r^{2}+20r\right)+\left(6r-15\right)$$

Factor out -4r in the first and 3 in the second group.

$$-4r\left(2r-5\right)+3\left(2r-5\right)$$

Factor out common term 2r-5 by using distributive property.

$$\left(2r-5\right)\left(-4r+3\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-8r^{2}+26r-15=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$r=\frac{-26±\sqrt{26^{2}-4\left(-8\right)\left(-15\right)}}{2\left(-8\right)}$$

Square 26.

$$r=\frac{-26±\sqrt{676-4\left(-8\right)\left(-15\right)}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$r=\frac{-26±\sqrt{676+32\left(-15\right)}}{2\left(-8\right)}$$

Multiply 32 times -15.

$$r=\frac{-26±\sqrt{676-480}}{2\left(-8\right)}$$

Add 676 to -480.

$$r=\frac{-26±\sqrt{196}}{2\left(-8\right)}$$

Take the square root of 196.

$$r=\frac{-26±14}{2\left(-8\right)}$$

Multiply 2 times -8.

$$r=\frac{-26±14}{-16}$$

Now solve the equation r=\frac{-26±14}{-16} when ± is plus. Add -26 to 14.

$$r=-\frac{12}{-16}$$

Reduce the fraction \frac{-12}{-16} to lowest terms by extracting and canceling out 4.

$$r=\frac{3}{4}$$

Now solve the equation r=\frac{-26±14}{-16} when ± is minus. Subtract 14 from -26.

$$r=-\frac{40}{-16}$$

Reduce the fraction \frac{-40}{-16} to lowest terms by extracting and canceling out 8.

$$r=\frac{5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{4} for x_{1} and \frac{5}{2} for x_{2}.

$$-8r^{2}+26r-15=-8\left(r-\frac{3}{4}\right)\left(r-\frac{5}{2}\right)$$

Subtract \frac{3}{4} from r by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-8r^{2}+26r-15=-8\times \frac{-4r+3}{-4}\left(r-\frac{5}{2}\right)$$

Subtract \frac{5}{2} from r by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-8r^{2}+26r-15=-8\times \frac{-4r+3}{-4}\times \frac{-2r+5}{-2}$$

Multiply \frac{-4r+3}{-4} times \frac{-2r+5}{-2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-8r^{2}+26r-15=-8\times \frac{\left(-4r+3\right)\left(-2r+5\right)}{-4\left(-2\right)}$$

Multiply -4 times -2.

$$-8r^{2}+26r-15=-8\times \frac{\left(-4r+3\right)\left(-2r+5\right)}{8}$$

Cancel out 8, the greatest common factor in -8 and 8.

$$-8r^{2}+26r-15=-\left(-4r+3\right)\left(-2r+5\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 -\frac{13}{4}x +\frac{15}{8} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{13}{4} rs = \frac{15}{8}$$

Two numbers r and s sum up to \frac{13}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{13}{4} = \frac{13}{8}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{13}{8} - u s = \frac{13}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{15}{8}

$$(\frac{13}{8} - u) (\frac{13}{8} + u) = \frac{15}{8}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{169}{64} - u^2 = \frac{15}{8}$$

Simplify the expression by subtracting \frac{169}{64} on both sides

$$-u^2 = \frac{15}{8}-\frac{169}{64} = -\frac{49}{64}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{49}{64} u = \pm\sqrt{\frac{49}{64}} = \pm \frac{7}{8}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{13}{8} - \frac{7}{8} = 0.750 s = \frac{13}{8} + \frac{7}{8} = 2.500$$

Evaluate

$$\left(3-4r\right)\left(2r-5\right)$$